Acids and Bases - conjugate pairs
Part I, Part IIAdvance discussion of acids bases
A second part is devoted to the subject of conjugation of acids and bases. The relationship between the acidic constant Ka, basic constant Kb, and the constant of autoionization of water, Kw will be discussed. The relationship is useful for weak acids and bases.
Skills to develop
Give three definitions for acids
Give three definitions for bases
Explain conjugate Acid-Base pair
Give the conjugate base of an acid
Give the conjugate acid of a base
Acids and Bases
The primitive way to characterize a substance is by tasting. Acids taste soure, and bases taste bitter.
At this point of your development, you already know some acids and bases. Are you familiar with the acids and bases listed here? If not, you probably will learn a lot here.
Acids and bases are also powerful concepts used to characterize substances. The concepts can and has been extended to much wider applications. For these reason, there are at least three definitions for each of acid and base. These definitions are given at different times to expand the concepts of acids and bases. This is why we do not even define acids and bases in this section.
Let us take a look how the concepts of acid and base are expanded over time. Acids and bases are related, and the relationship is called conjugation. An acid has a conjugate base and vice versa.
Evolution of the Acid-Base Concept
In 1884, Arrhenius noticed that all acids have H+ ions and bases have OH- ions. Thus, he considered all substances giving H+ and OH- ions are acids and bases respectively.
Acid Formula Acetic CH3COOH Hydrochloric HCl
Sulfuric H2SO4 Nitric HNO3 Carbonic H2CO3
Base Formula Ammonia
water NH4OH Sodium
hydroxide NaOH sodium
Less than 40 years later, in 1923, Bronsted, Lowry and Lewis wanted to expand the concepts of acids and bases so that a wider area of
chemistry can be understood using the same principle.
Bronsted-Lowry's concept focused on the proton, and defined
Acids as proton donors,
Bases as proton acceptors,
whereas Lewis focused on the electrons and defined
Acids as electronphils,
Bases as nucleophils.
Lewis definitions are more general than Bronsted-Lowry's definitions, they are now widely accepted and practiced. The more general definitions for acids and bases allow us to discuss more chemical reactions as acid-base reactions.
Conjugate Acid-Base Pairs
According to the Bronsted-Lowry theory of acids and bases, an acid is a proton donor and a base is a proton acceptor. Once, an acid has given up a proton, the remaining part can be a proton acceptor, and thus a base. In this regard, an acid and a base are closely related to one another.
H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of Acid
-For example:
NH3 + H2O = NH4+ + OH
-HAc = H+ + Ac
-Thus, NH4+ and NH3 are a pair of conjugate acids and bases, as are HAc and Ac-. The table lists conjugate acid-base pairs for your reference so that you can figure out the strategy of identifying them. Furthermore, for an acid or base, you should be able to give its conjugate base or acid.
Confidence Building Questions
What is the conjugate base of HF? Answer... F
-Consider... HF = H+ + F
Acid Base Neutraliztion
1884 Arrhenius
ionize
H+ ionizeOH- H
+ + OH --> H2O 1923 Bronsted-Lowry Proton donor Proton acceptor
HA + B -> A + HB 1923
Lewis Electrophil Nucleophil E + :Nu -> E:Nu
Conjugate acid
Conjugate base H3O+H2O
H2O OH -H2SO4HSO4 -HSO4-SO4
2-NH4+NH3 NH3NH2 -H3PO4H2PO4 -H2PO4-HPO4 2-HPO42-PO4 3-CH3COOH CH3COO
What is the conjugate base of benzoic acid, C6H5COOH?
Answer... C6H5COO
-Consider... HC6H5COO- -> H+ + C6H5COO
- What is the conjugate acid of benzoate, C6H5COO-?
Answer... C6H5COOH
Consider... Recall the previous question.
What is the conjugate base of ammonium ion NH4+?
Answer... NH3
Consider... NH4+ = NH3 + H+
What is the conjugate acid of sulfate ion SO42-?
Answer... HSO4
-Consider... This is the conjugate base of H2SO4; HSO4- can be either an acid or a base.
What is the conjugate acid of HSO4
-Answer... H2SO4
Consider... H2SO4 has no charge.
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Strong Acids and Bases
Skills to develop
Give the names and formulas of some strong acids and bases.
Explain the pH scale, and convert pH and [HSUP>+].
Evaluate solution pH and pOH of strong acids or bases.
Strong Acids and Bases
The animation here shows the formation of H3+O ions and OH- ions in an aqueous system.
Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life.
The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, [H+].
Strong acids are acids that are completely or nearly 100% ionized in their solutions. Here are some common strong acids: Ionization of a strong acid HA can be represented by:
HA = H+ + A x x
where x is the concentration of H+, [H+]. For a strong
acid, [H+] = [A-] = concentration of acid (= x), if x is much greater than 1x10-7
(represented as e-7). For a very dilute strong acid solution with concentration less than 1E-7, the pH is dominated by the autoionization of water,
H2O = H+ + OH-, Kw = 1e-14 at 298 K.
The pH and pOH Scales
The pH scale is defined as the negative log of the concentration of H+:
pH = -log[H+]
The pOH scale is defined as the negative log of the concentration of OH-, [OH-]:
pOH = -log[OH-]
With this scale, calculating the pOH can be done in the same manner as the pH scale. Example 1
Calculate the pH of a solution with 1.2345E-4 M HCl
Solution
The solution of a strong acid is completely ionized. Thus, [H+] = 1.234e-4. pH = -log(1.234e-4) = 3.909
Discussion
What is the pH for a solution containing 1.234 M [HCl]? pH = 0.0913 Example 2
Calculate the pH of a stock HCl solution that is 32% by mass HCl.
Solution
The density of such a solution is needed before we can calculate the pH. Since the density is not on the label, we need to find it from the Material Safety Data Sheet, which gives the specific gravity of 1.150. Thus, the amount of acid in 1.0 L is 1150 g.
The amount of HCl = 1000*1.150*0.32
= 368 g (1 mol/36.5 g <- molar mass of HCl)
Strong acids
Type Formula
Hydrogen
halides HCl HBr HI Oxyacids
of halogens,
HClO3 HClO4
HBrO3 HBrO4
HIO3 HIO4 Sulfuric acid H2SO4
Nitric acid HNO3
The pH and pOH scale at 298 K pH [H+] pOH
= 10.08 M = [H+]
pH = -log(10..08) = -1.003
Discussion
Yes, pH have negative values if [H+] > 1.0
Check out the information on nitric acid, and calculate the pH of a stock nitric acid solution.
Example 3
Calculate the pH of a solution containing 1.00E-7 of HCl.
Solution
[H+] = 1.0e-7 M from the strong acid, and if x is the amount from the ionization of water, then we have the equilibrium due to the autoionization of water:
HCl = H+ + Cl
1E-7 1E-7 <--- [H+] from the acid
H2O = H+ + OH
(1E-7)+x x <--- we don't know yet
Recall that Kw = [H+] [OH-] = 1E-14, due to the ionization equilibrium of water in the solution:
{(1.00E-7)+x} x = 1E-14 x2 + 1.00e-7x - 1.00E-14 = 0 Solving this equation for x results in
x = {-1.00E-7 + (1.00E-14 + 4*1.00E-14)1/2} / 2 = 0.61E-7
[H+] = (1.00 + 0.61)E-7 M pH = -log(1.61E-7) = 6.79
Discussion
If you require only 1 significant figure, the pH is about 7.
Strong Bases
Strong bases are completely ionized in solution. For example, KOH dissolve in water in the reaction
KOH = K+ + OH-.
Relative to strong acids, there are fewer number of strong bases. Most strong bases are alkali hydroxides. Calcium oxide is considered a strong base, because it is completely, almost
Strong bases Name Formula Sodium
hydroxide NaOH Potassium
hydroxide KOH Cesium
hydroxide CsOH Calcium
completely, ionized. However, the solubility of calcium hydroxide is very low. When Ca(OH)2 dissolve in water, the ionization reaction is as follows:
Ca(OH)2 = Ca2+ + 2 OH-.
The concentration of OH- is twice the concentration of Ca2+, [OH-] = 2 [Ca2+]
Example 4
Calculate the pOH of a solution containing 1.2345E-4 M Ca(OH)2.
Solution
Based on the ionization,
[OH-] = 2*1.234e4 = 2.468E-4 M pOH = -log(2.468E-4)
= 3.61
Discussion
The molar solubility of calcium hydroxide is 0.013 M Ca(OH)2. Calculate the pOH. pOH = 1.58
Autoionization of Water
The equilibrium product Kw = [H+] [OH-] is a constant at a definite temperature due to the autoionization of water,
H2O = H+ + OH-.
At 298 K, Kw = 10-14 and the following relationship in any aqueous solution is obvious, pOH + pH = 14 at 298 K.
Of course, when T is higher than 298 K, pH + pOH is slightly less than 14 due to a higher degree of ionization of water. Conversely, at low temperatures, pH + pOH is larger than 14 due to less degree of ionization.
Confidence Building Questions
What is the pH of a solution containing 0.01 M HNO3?
Answer 2
Hint...
You do not need a calculator to evaluate -log (0.01) = 2
What is the pH of a solution containing 0.0220 M Ba(OH)2? Give 3
Answer 12.64
Hint...
Ba(OH)2 = Ba2+ + 2 OH
- Exactly 1.00 L solution was made by dissolving 0.80 g of NaOH in water. What is [H+]? (Atomic mass: Na, 23.0; O, 16.0; H, 1.0)
Answer 5.0e-13
Hint...
[OH-] = 0.80/40 = 0.02 M; [H+] = 1e-14 / 0.02 = 5e-13 M. The pH is -12.3.
What is the pH for a solution which is 0.050 M HCl? Answer 1.3
Hint...
This solution contains 1.83 g of HCl per liter. [H+] = 0.050.
Which of the following is usually referred to as strong acid in water solution? HF, HNO2, H2CO3, H2S, HSO4-, Cl-, HNO3, HCN
Answer HNO3 Consider...
All others are weak acids
he pH scale
Skills to be tested
Discuss the pH scale
Point out the neat things about the pH scale.
Tell the origin and the logic of of using the pH scale.
Apply the same strategy for representing other types of quantities such as
pKa, pKb, pKw.
The pH scale
From the simple definition of pH being the negative log of the H+ ion concentration, [H+], pH = - log [H+]
This scale is convenient to use, because it converts some odd expressions such as 1.23x10-4 into a single number of 3.91.
This scale covers a very large range of [H+], from 0.1 to 10-14. When [H+] is high, we usually do not use the pH value, but simply the [H+]. For example, when [H+] = 1.0, pH = 0. Because we seldom say the pH is 0, and that is why you consider pH = 0 such an odd expression. A pH = -0.30 is equivalent to a [H+] of 2.0 M. Negative pH values are only for academic exercises. Using the concentrations directly conveys a better sense than the pH scales.
The pH scale expands the division between zero and 1 in a linear scale or a compact scale into a large scale for comparison purpose. In mathematics, you learned that there are infinit values between 0 and 1, or between 0 and 0.1, or between 0 and 0.01 or between 0 and any small value. Using a log scale certainly converts infinite small quantities into infinite large quantities.
The non-linearity of the pH scale in terms of [H+] is easily illustrated by looking at the corresponding values for pH between 0.1 and 0.9 as follows:
pH = 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 [H+] = 0.79 0.63 0.50 0.40 0.32 0.25 0.20 0.16 0.13
Because the negative log of [H+] is used in the pH scale, the pH scale usually have positive values. Furthermore, the larger the pH, the smaller the [H+].
Other Interesting Facts about the pH Scale
According to the Oxford Dictionary, the pH scale was originally introduced by the Danish biochemist S.P.L. Sorensen in 1909 using the symbol pH. Other symbols such as pH have been used in the past. The letter p is derived from the German word potenz meaning power or exponent of, in this case, 10.
You may argue that what have we learned by looking into historical origin of the term? Well, the origin of concept is interesting in that we sometimes need to develop concepts ourselves. A concept or tool becomes important if many people find it convenient and elegant.
The reality is that many chemists have used the pH scale and the p scales for many other quantities that they often take it for granted, without realizing the logic behind their usages. For example, we have used the pKa, pKb, pKw, notations by analogy to the pH notations with out asking a question.
Now that you know the pH is an exponent, the following relationship is obvious: [H+] = 10-pH
What is the [H+] in a solution whose pH is 3.21?
Answer 6.17e-4
Consider... [H+] = 10-pH
Weak Acids and Bases
Skills to develop
Define a weak acid or base.
Calculate pH and pOH of a weak acid or base solution using simple
formula, quadratic equation, and including autoionization of water.
Calculate the pH or pOH quickly.
Weak Acids and Bases
Weak acids and bases are only partially ionized in their solutions, whereas strong acids and bases are
completely ionized when dissolve in water.
Some common weak acids and bases are given here. Furthermore, weak acids and bases are very
common, and we encounter them often both in the academic problems and in everyday life.
The ionization of weak acids and bases is a chemical equilibrium phenomenon. The equilibrium principles are essential for the understanding of equilibria of weak acids and weak bases.
The conjugate acid-base pairs have been discussed in Acids and Bases. In this connection, you probably realize that conjugate acids of weak bases are weak acids and conjugate bases of weak acids are weak bases.
Ionization of Week Acids
Common Weak Acids
Acid Formula
Formic HCOOH Acetic CH3COOH Trichloroacetic CCl3COOH
Hydrofluoric HF Hydrocyanic HCN
Hydrogen sulfideH2S
Water H2O Conjugate acids
of weak basesNH4+
Common Weak Bases Base Formula ammonia NH3 trimethyl
ammoniaN(CH3)3 pyridine C5H5N ammonium
hydroxideNH4OH water H2O HS- ion HS -conjugate bases
-Acetic acid, CH3COOH, is a typical weak acid, and it is the
ingradient of vinegar. It is partially ionized in its solution. CH3COOH = CH3COO- + H+
The structure of the acetate ion, CH3COO-, is shown on the
right.
Example 1
In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.
Solution
From the ionization of acetic acid,
CH3COOH = CH3COO- + H+
0.100 0.0042 0.0042 we conclude that
[H+] = [CH
3COO-]
= 0.0042.
Thus, pH = -log0.0042 = 2.376.
The equilibrium constant of ionzation,
(0.0042)2
K = --- = 1.78x10-5
1.000
Discussion
The equilibrium constant of an acid is represented by Ka; and similar to the pH scale, a pKa scale is defined by
pKa = - log Ka
and for acetic acid, pKa = 4.75. Note that Ka = 10-pKa
The pH and p
K
a of Weak acid
There are many weak acids, which do not completely dissociate in aqueous solution. As a general discussion of weak acids, let HA represent a typical weak acid. Then its ionization can be written as:
HA = H+ + A
-In a solution whose label concentration is C (= [HA] + [A-]), let us assume that x is
the concentration that has undergone ionization. Thus, at equilibrium, the concentrations are
H O | //
O-[HA] = C - x
[H+] = [A-] = x
Make sure you understand why they are so, because you will have to setup these relationship in your problem solving. In summary, we formulate them as
HA = H+ + A
-C - initial concentration, assume x M ionized
C-x x x - equilibrium concentration
x2 Ka = C - x
pKa = - log Ka
The pKa values of many weak acids are listed in table form in handbooks, and some of these values are given in the Handbook of CAcT, p a K of Acids.
Example 2
The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentrations of 1.0 M, 0.010 M, and 0.00010 M.
Solution
Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the formulation below.
CH3COOH = CH3COO- + H+
C - x x x
x2
Ka = C - x
The equation is then
x2 + Ka x = C Ka = 0
The solution of x is then
- Ka + (Ka2 + 4 C Ka)(1/2) x = 2
Recall that Ka - 1.78e-5, the values of x for various
C are given below:
C = 1.0 0.010 0.00010 M x = 0.0042 0.00041 0.0000342 M pH = 2.38 3.39 4.47
Discussion
In the above calculations, the following cases may be considered:
1.
2. If x is small (< 1% of) compared to C, then C-x is 3. approximately C. Thus,
4.
5. x = (Ka * C)(1/2) 6.
7. Note that you are comparing x with C here. If C > 8. 100*Ka, the above method gives satisfactory results. 9.
10.
11. If x is not small by comparison with C, or C is not large 12. in comparison to Ka, then the equation takes this form:
13.
14. x2 + Ka x - C Ka = 0, 15.
16. and the solution for x, which must not be negative, has been 17. given above.
18. 19.
20. Both cases 1 and 2 neglect the contribution of [H+] from the ionization 21. of water. However, if the pH calculated from cases 1 and 2 falls in the 22. range between 6 and 7, the concentration from self-ionization of water 23. cannot be neglected.
24.
25. When the contribution of pH due to self-ionization of water cannot be 26. neglected, there are two equilibria to be considered.
27. HA = H+ + A -28. C-x x x
29.
30. H2O = H+ + OH
-31. 55.6 y y <- - - ([H2O] = 55.6) 32.
33. Thus, [H+] = (x+y), 34. [A-] = x, 35. [OH-] = y, 36.
37. and the two equilibria are 38.
39. (x+y) x
40. Ka = --- ... (1) 41. C - x
42.
43. and
44. Kw = (x+y) y, ... (2) 45. (Kw = 1E-14)
46.
47. There are two unknown quantities, x and y in two equations, and (1) may 48. be rearranged to give
50. x2 + (y + Ka) x - C Ka = 0 51.
52. -(y+Ka) + ((y+Ka)2 + 4 C Ka)(1/2)
53. x = ---54. 2
55. 56.
57. One of the many methods to find a suitable solution for this problem is 58. to use iterations, or successive approximations.
59.
1. Assume that y = 1E-7 2.
3.
4. Calculate an x value using the quadratic form
5. -(y+Ka) + ((y+Ka)2 + 4 CKa)(1/2)
6. x = ---7. 2
8. 9.
10. Calculate a new y value (yn) from the x just obtained using 11.
12. yn = 1E-14/(x+y) 13.
14.
15. Replace y in step (2) by yn, and recalculate x. 16.
17.
18. Repeat steps (2) and (3) until the new values and the 19. old values differ insignificantly.
20. 60.
The above procedure is actually a general method that always gives a satisfactory solution. This technique have to be used to calculate the pH of dilute weak acid solutions. Further discussion is given in the
Exact Calculation of pH.
Calculate pOH of Basic Solutions
The discussion on weak acids provide a paradigm for the discussion of weak bases. For weak base B, the ionization is
B- + H
2O = HB + OH
-and
[HB] [OH-]
Kb =
[B-]
The pOH can be calculated for a basic solution if Kb is given. In this case, the
discussion is similar and parallel to that given above for the calculation of pH of weak acids when Ka is know.
A weak acid is a compound that a. is completely ionized in solution, b. is not completely ionized in solution, c. gives a high pH in a solution,
d. gives a low pH in its solution,
Answer b
Consider...
The pH of a solution depends on both the concentration and the degree of ionization, (or using Ka as an indicator). In contrast, a strong acid is
completely ionized in solution.
The acidity constant, Ka, for a strong acid is a. infinity,
b. very large, c. very small, d. zero.
Answer b
Consider...
Infinity is a concept, it does not represent a definite value. Derive your answer from the definition of equilibrium constant. A strong acid is
"completely" ionized in its solution, but the concentration of the conjugate acid is not zero. Thus, a very large Ka is more realistic than infinity.
Household vinegar is usually 5% acetic acid by volume. Calculate the molarity of this solution.
Assume density of solution to be 1 g/mL. The formula weight of CH3COOH is 60.
Answer 0.8 M
Consider...
Assume 1 L solution. You have 50 mL acetic acid in 1 L vinegar. The density is 1 g/mL, thus, you have 50 g acetic acid. There is 50 mL vinegar in 1.0 L of vinegar, 50 g/60 g per mol = 0.8 mol/L
Acetic acid is a typical and familiar compound that provides a good example for numerical problems. Its acidity constant Ka is 1.85e-5. What is the pH of a concentrated vinegar, which is a 1.0 M acetic acid solution?
Answer 4.3e-3
Consider...
Use the approximation of [H+] = square root of (K
a * C). [H+] = (1.85e-5)1/2 =
that most text books give Ka = 1.75e-5, but we assume a slightly different value in this and the following problems.
If you dilute the vinegar 100 times in a soup that you are cooking, the concentration of your soup is 0.010 M in acetic acid.
Other ingredients are ignored. What is the pH of this solution?
Answer 3.4
Consider...
Use the approximation of [H+] = sqrt (K
a * C). The concentration of H+ goes
from 3.3E-3 in a 1 M solution down to 4.3E-4 M in a 0.01 M solution. The concentration of H+ decreases 10 times when the concentration of the acid
decreases 100 times.
What is the pH of a 0.010 M HCl solution? Answer 2
Consider...
[H+] = 0.010 M. The concentration of H+ is from HCl, which is a strong acid.
The pH of a 0.01 M HCl solution is lower than that of a 1 M acetic acid solution, compare with the previous problem.
What is the pH of a 1.0E-4 acetic acid solution (Ka = 1.85E-5)? Answer 4.4
Hint...
Note that [H+] = 4.3E-5 is 4% of [HAc] (= 1.0E-4). Use the formula
-Ka + (Ka2 + 4 C Ka)(1/2) x = 2
and see what you get. You should use the quadratic formula to calculate [H+]. The value using the quadratic formula is 4.5 rather than 4.4 from
sqrt(C*Ka)
What is the pH of a 1.0E-3 M chloroacetic acid solution (Ka = 1.4E-3)? This is an interesting numerical problem. Make a good effort to solve it.
Answer 3.2
Hint...
Even a strong acid with concentration of 1.0E-3 M gives a pH of 3. When C and Ka are comparable, you have to use the quadratic formula.
Answer 6.0
Consider...
At this concentration, the acid is almost completely ionized.
What is the pH of a 1.0e-7 M chloroacetic acid solution (Ka = 1.4e-3)? Answer 6.7
Consider...
From the previous question, you know that the chloroacetic acid should have been completely ionized. Thus, [H+] is about 2e-7, half of that is
contributed by the self-ionization of water. This corresponds to a pH of 6.7.
You have done a number of numerical problems involving various concentrations of some weak acids. These problems are inter- related. If you do not yet have the complete picture, you should review all these questions. Better yet, review the module. What is the pH of a 1.0E-9 M solution of chloroacetic acid, Ka = 1.4E-3?
Answer 7
Consider...
Critical judgment is required. The pH is entirely due to the self-ionization of water at this concentration.
Conjugate Acids of Bases - K
a
K
b
and K
w
Part I, Part II
Advance discussion of acids bases
Skills to develop
Explain conjugate acids of bases.
Evaluate Ka of the conjugate acid of a base.
Treat the conjugate acid of a base as an acid in numerical calculations.
Reverse the role of acid and base for the previous skills.
Conjugate Acids of Bases
The conjugation of acids and bases have been discussed earlier. After losing a proton, the
H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of Acid
-For example:
NH3 + H2O = NH4+ + OH
-HAc = H+ + Ac
-Thus, NH4+ and NH3 are a pair of conjugate acids and bases, as are HAc and Ac-.
K
aValues of Conjugate Acids of Bases
We have used Ka and Kb as the acidic and basic constants of acids and bases. Can an acidic constant, Ka, be assigned to the conjugate acid of a base? If so, what is the
relationship between Ka of the conjugate acid and Kb of the base? We are going to derive the relationship here. Note that water always plays a role in the conjugation acid-base pair.
Let BH+ be the conjugate acid of a base, then the expression for the acidic constant Ka for the conjugate acid:
BH+ = B + H+ can be written as [B] [H+] Ka = [BH+]
[B] [H+] [OH-] = --- [BH+] [OH-]
[B]
= --- [H+] [OH-] [BH+] [OH-]
1 = --- Kw Kb
Thus,
KaKb = Kw
Furthermore,
- log (Ka) - log (Kb) = -log (Kw)
and at 298 K, we have pKa + pKb = 14. Examples 1
Solution
The conjugate base is CO32-. Kb = (1E-14)/(4.7E-11) = 2.1E-4
DIscussion
The Kb so calculated is for the reaction, CO32- + H2O = HCO3- + OH
[HCO3-] [OH-] Kb = [CO32-]
The anion CO32- is a rather strong base, and the large value calculated for Kb agrees with the fact.
Example 2
The Kb for the anion of oxalic acid, COO- | COOH is 1.8E-10. What is Ka for
the oxalic acid (COOH)2?
Solution
The Ka for oxalic acid is Ka = (1E-14) / (1.8E-10) = 5.6E-5
DIscussion
The calculation regarding Ka and Kb conversion is simple, but understanding what problems require this type of conversion is difficult. The concept is rather useful, and it further broadens the concept of acid and base.
K
bValues of Conjugate Bases of Acids
We can also calculate the Kb value of the conjugate base from the Ka value of its conjugate acid. The principle is the same as that used to calculate the Ka values of the conjugate acid of a base as we have just discussed.
Let A- be the conjugate base of an acid HA, then the expression for the equilibrium constant for the reaction:
A- + H2O = HA + OH -can be written as
Kb = [A-]
Multiplying the numerator and denominator with [H+] leads to, [HA] [OH-] [H+]
Kb = -- [A-] [H+]
Rearrangement gives
[HA]
Kb = --- [OH-] [H+]
[A-] [H+]
[HA]
= --- Kw
[A-] [H+]
Kw
=
Ka
Thus,
Ka Kb = Kw
and this formula is the same as the one derived for the conjugate acid of a base. Again, at 298 K, we have
Ka Kb = 1E-14
and the value for Kw is larger than 1E-14 at higher temperatures. Kw is smaller at temperature less than 298 K.
The concept of conjugate acid and base pairs are very useful for the consideration of acidity and basicity of salts. The applications of the relationship,
Ka Kb = Kw
are further illustrated on the topic of Hydrolysis. Hydrolysis reactions are reactions of cations or anions of salts with water. As a result of these reactions, a salt solution is either acidic or basic.
Confidence Building Questions
Calculate Kb for the acetate ion from the Ka for acetic acid of 1.8E-5.
Answer 5.6E-10 Consider...
Kb = (1e-14)/(1.8e-5) = 5.6E-10 If Kb for the acetate ion is 5.6E-10, what is Ka for acetic acid?
The Ka for trimethylammonium ion (CH3)3NH+ is 1.6E-10. Calculate Kb for its
conjugate base. Answer 6.25E-5 Consider...
Ka = (1e-14)/(1.6e-10) = 6.25E-5 You know all about conjugate acid-base pairs now. Learning is a pleasure.
At some temperature, Kw = 1e-13. Calculate the Kb value for the acetate ion.
(Ka for acetic acid is 9.5E-5 at the same temperature).
Answer 1.05e-9 Consider...
Kb = 1e-13/9.5e-5 The acidic constants are dependent on temperature.
© cchieh@
Exact pH calculation
Skills to develop
Calculate the pH when two weak acids are present in a solution.
Calculate the pH when the concentration of the acid is very dilute.
Calculate the pH by including the autoionization of water.
Proof read and polish is still needed for this module.
The paradigm given in Weak Acids and Bases deals with the equilibria of solutions containing one weak acid or one weak base. In most cases, the amount of H+ from the autoionization of water is negligible. For very dilute solutions, the amount of H+ ions from the autoionization of water must also be taken into account. Thus, a strategy is given here to deal with these systems.
When two or more acids are present in a solution, the concentration of H+ (or pH) of the solution depends on the concentrations of the acids and their acidic constants Ka. The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, Ka's.
Similarly, the concentration of OH- ions in a solution containing two or more weak bases depends on the concentrations and Kb values of the bases.
For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases.
Two Equilibria with a Common Ion H
+If the pH is between 6 and 8, the contribution due to autoionization of water to [H+] should also be considered. When autoionization of water is considered, the method is called the exact pH calculation or the exact treatment. This method is illustrated in this module. The need for the eaxct treatment has been pointed out in the discussion of Example 2 in Weak Acids and Bases.
When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to consider:
HA = H+ + A C-x x x
H2O = H+ + OH
55.6 y y - - - ([H2O] = 55.6)
Thus,
[H+] = (x+y),
[A-] = x,
[OH-] = y,
and the two equilibrium constants are
(x + y) x
Ka = --- - - (1) C - x
Kw = (x + y) y, - - - (2)
Although you may use the method of successive approximation, but the formula to calculate the pH can be derived.
Solving for x from (2) gives Kw
x = ---- - y y
and substituting this expression into in (1) results in
(x+y) (Kw/y - y) Ka = C - Kw/y + y
Rearrange this equation to give:
[H+] = (x+y)
C - Kw/y + y = --- Ka Kw/y - y
Note that Kw/y = [H+], and y = Kw/[H+]. Thus, we get:
C - [H+] + Kw/[H+]
[H+] = --- Ka [H+] - Kw/[H+]
To simplify the formula, let
h = [H+]
Then, the formula becomes
C - h + Kw/h h = --- Ka h - Kw/h
Now, consider the approximations we can make when we use this formula due to conditions.
Case 1
C - h
h = --- Ka h
h2 + Ka h - C Ka = 0
and
-Ka + (Ka2 + 4 C Ka)(1/2)
h =
2
Case 2
If h << C, then C - h => C C
h = --- Ka h
h = (C Ka)1/2
The treatment presented as Case 3 at the beginning of this tutorial is more general, and may be applied to problems involving two or more weak acids in one solution.
Example 1
Calculate the [H+], [Ac-], and [Cc-] when the solution contains 0.200 M HAc
(Ka = 1.8E-5), and 0.100 M HCc (the acidity constant Kc = 1.4E-3). (HAc is
acetic acid whereas HCc is Chloroacetic acid). Solution
Assume x and y to be the concentrations of Ac- and Cc-, respectively, and write the concentrations below the equations:
HAc = H+ + Ac 0.200-x x x
HCc = H+ + Cc 0.100-y y y
[H+] = (x + y)
(x + y) x
--- = 1.8E-5 - - - (1) 0.200 - x
(x + y) y
--- = 1.4E-3 - - - (2) 0.100 - y
Solving for x and y from (1) and (2) may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since HAc is a weaker acid than is HCc, you expect x << y. Further, y << 0.100. Therefore, x + y => y and 0.100 - y => 0.100. Equation 2 becomes:
( y) y
--- = 1.4E-3 - - - (2') 0.100
which leads to
y = ((1.4E-3)(0.100))1/2 = 0.012
Substituting y in (1) results in
(x + 0.012) x
--- = 1.8E-5 - - - (1') 0.200 - x
This equation is easily solved, but you may further assume that 0.200 - x => 0.200, since x << 0.200. Thus,
-0.012 + (1.44E-4 + 1.44E-5)1/2 x = 2
= 2.9E-4 - - - - Small indeed compared to 0.200
You had a value of 0.012 for y by neglecting the value of x in (2). You can now recalculate the value for y by substituting values for x and y in (2).
(2.9E-4 + y) y
--- = 1.4E-3 - - - (2") 0.100 - 0.012
Solving for y in the above equation gives
y = 0.011.
You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from 2.9E-4 to a new value of 3.2E-4. Use your calculator to obtain these values.
Discussion
You should write down these calculation on your note pad, since reading alone does not lead to thorough understanding.
Example 2
A weak acid HA has a Ka value of 4.0E-11. What are the pH and the
equilibrium concentration of A- in a solution of 0.0010 M HA?
Solution
For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a).
Method (a) The two equilibrium equations are: HA = H+ + A-;
0.0010-x x x
H2O = H+ + OH y y
[H+] = (x+y)
(x+y) x
--- = 4.0E-11 - - (3); 0.0010-x
(x+y) y = 1e-14 - - - (4)
Assume y << x, and x << 0.0010, then you have
(x ) x
--- = 4.0E-11 - - - (3') 0.0010
x = ((0.0010)(4.0e-11))1/2 = 2.0E-7
Substituting 2.0E-7 for x in 4 and solve the quadratic equation for y gives, (2.0E-7+y) y = 1E--14
y = 4.1E-8
Substituting 4.1E-8 in (3), but still approximate 0.0010-x by 0.0010 (x+4.1E-8) x
--- = 4.0E-11 - - - (3'') 0.0010
Solving this quadratic equation for a positive root results in x = 1.8E-7 M <--- Recall x = [A-]
[H+] = x + y
= (1.8 + 0.41)1E-7 = 2.2E-7
pH = 6.65
Method (b)Using the formula from the exact treatment, and using 2E-7 for all the [H+] values on the right hand side, you obtain a new value of [H+] on the left hand side,
C - [H+] + Kw/[H+]
[H+] = --- Ka [H+] - Kw/[H+]
= 2.24E-7
pH = 6.65.
The new [H+] enables you to recalculate [A-] from the formula: (2.24E-7) [A-] = C Ka
[A-] = (0.0010) (4.0E-11) / (2.24E-7) = 1.8E-7
Discussion
Often, you may be attempted to use the approximation method: x = (C Ka)1/2
= 2.0E-7 M A-, or H+; pH = 6.70 and obtained a pH of 6.70.
The value 6.70 is greater than 6.65 by less than 1 %. However, when an approximation is made, you have no confidence in the calculated pH of 6.70.
Calculating pH by Approximations
Water is both an acid and a base due to the autoionization, H2O = H+ + OH
-However, the amount of H+ ions from water may be very small compared to the amount from an acid if the concentration of the acid is high.
When calculating [H+] in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids.
Confidence Building Questions
A solution contains 0.200 M HAc (Ka = 1.8E-5), and 0.100 M HCc (the acidity
constant Kc = 1.4E-3). (HAc is acetic acid whereas HCc is Chloroacetic acid). Assume x and y are the concentrations of Ac- and Cc-, respectively, then
HAc = H+ + Ac- HCc = H+ + Cc
- 0.200-x x x 0.100-y y y
Which is larger, x or y?
Answery
Consider...
Which is the stronger acid of the two? Most of the H+ will be from the stronger acid. Since HCc is a stronger acid (Kc =78 Ka), acidity is dominated by the ionization of HAc. Approximation to be made is to ignore x, i.e. x + y => y
Calculate the pH of a solution containing 1.0E-3 M chloroacetic acid (Ka =
1.4E-3), and 3.0E-3 acetic acid (Ka = 1.8E-5).
Answer 3
Consider...
You know that the pH is dominated by the stronger acid of the two. Based on that, the pH is 2.9. Can approximation be made?
Calculate the pH of a solution containing 1.0E-6 M chloroacetic acid, (Ka =
1.4E-3). Don't forget that water is a weak acid here, (Kw = 1E-14).
Answer 6
Consider...
At this concentration, the chloroacetic acid is completely ionized. The contribution from the self-ionization of water is still unimportant.
AcidBase: Fraction of Dissociation
Skills to develop
Define the fraction of dissociation of a weak electrolyte
Calculate the fraction of dissociation of a weak acid or base
Sketch the fraction of dissociation as a function of concentration
Fraction of Dissociation
A chemical equilibrium involving dissociation can be represented by the following reaction.
AB = A + B, - - - K = [A][B]/[AB]
The concept equilibrium has been discussed in Mass action law and further discussed in
Weak Acids and Bases, Ka, K b, and K w, and Exact pH Calculations.
has reached equilibrium. The concentration of A and B will each then be f C. Note that C is also the total concentration.
For convenience, we can write the concentration below the formula as: AB = A + B
(1-f)C fC fC - - - Concentration
and the equilibrium constant, K = [A][B]/[AB] can be written as (using the concentration below the formula):
f2 C K = 1 - f
Variation of
f
as a Function of
C
How does f vary as a function of C? Common sense tells us that f has a value between 0 and 1 (0 < f < 1). For dilute solutions, f => 1, and for concentrate solutions, f => 0. In solving the above equation for f, we obtain:
- K + (K2 + 4 K C)1/2 f = 2 C
Note that f is the fraction of molecules that have dissociated. It is also called the degree of ionization. When converted to percentage, the term percent ionization is used. Even with the given formulation, it is still difficult to see how f varies as C changes. In the DOS version CACT, there is a program associated with this module for you to investigate the relationship between f and C. The program plots f as a function of C for a specified value of K. We now, illustrate the variation with a table below for a moderate value of K = 1.0e-5.
Variation of fraction of dissociation f as a function of concentration C when K = 1.0e-5.
C 1e-7 1e-6 1e-5 1e-4 1e-3 0.01 0.1 1.0 10 100
f 0.99 0.92 0.62 0.27 .095 .031 1e-2 3e-3 1e-3 3e-4
There is little change in f when C decrease from 1.0e-7 to 1.0e-6, but the changes are rather some what regular for every 10 fold decrease in concentration. Please plot f against a log scale of C to see the shape of the variation as your activity. Normally, we will not encounter solution as dilute as C = 1.0e-7, and we will never encounter solution as concentrate as 100 M either.
What is the fraction of dissociation for a strong acid?
Answer 1
Consider...
A strong acid is almost completely dissociated.
What is the fraction of dissociation for a compound that does not dissociate.
Answer 0
Consider...
Since there is no dissociation, the fraction is zero. This is redundant question.
A 0.1 M solution of an acid HB has half of its molecules dissociated. Calculate the acidity constant Ka.
Answer 0.05
Consider...
HB = H+ + B
-0.05 -0.05 -0.05 M <-- Concentrations at equilibrium.
K = ?
The equilibrium constant for a weak base B is 0.05, what is the fraction of dissociation if the concentration is 0.10 M?
Answer 0.5
Consider...
See the previous question.
The equilibrium constant for a weak base B is 1.0e-3, what is the fraction of dissociation if the concentration is 0.10 M?
Answer 0.095
Consider...
Make a table to see the variation of f when the concentration changes from 1e-7 to 100 in steps of 10 folds as given previously.
AcidBase: Polyprotic Acids
Skills to develop
Define a polyprotic acid.
Write the equilibrium equations of ionization of polyprotic acids. Calculate acidity constants, K1, K2, K3, and the overall K.
Calculate the concentrations of various species for a given set of data
Proof reading and polish still required!
Polyprotic Acids
Some polyprotic acids are given in the table on the right here. Knowing their names and familar with their properties (ionization for example) is an asset for you. Polyprotic acids contain more than one mole ionizable hydronium ions per mole of acids. They ionize to give more than one H+ ions per molecule. Possible forms of
three polyprotic acids are given below after their dissociation into H+ ions. H2S, HS-, S
H2SO4, HSO4-, SO4
H3PO4, H2PO4-, HPO42-, PO4
3-These acids ionize in several stages, giving out one proton at each stage. The acidity constants for these acids may be written as K1, K2, K3, ...
Consider H2S,
H2S = H+ + HS
[H+] [HS-] K1 = [H2S] and
HS- = H+ + S
[H+] [S2-] K2 = [HS-]
Obviously, for the overall ionization reaction, H2S = 2 H+ + S
[H+]2 [S2-] Koverall = [H2S]
= K1 K2
Confirm the above obvious result on a sheet of paper to satisfy yourself. For polyprotic acids, the following is always true:
Some Polyprotic Acids
Formula Name
K1 > K2 > K3 > ...
For most acids, K1/K2 = 1E5 or 100000, and K2/K3 = 1E5, but oxalic acid is different. For oxalic acid, K1 = 5.6E-2, and K2 = 5.4E-4. The two acidic groups are separated by a C-C bond in oxalic acid.
Examples 1
Calculate the overall equilibrium constant for oxalic acid H2C2O4 = 2 H+ + C2O42-.
K1 = 5.6E-2
K2 = 5.4E-5
Solution
The calculation is straight forward: Koverall = K1 K2
= 3.0E-6 Example 2
A solution is acidified with HCl so that its pH is 1.0, and is saturated with H2S at 298 K. What is the sulfide S2- ion concentration in this solution? At
298 K, a saturated H2S solution has [H2S] = 0.10 M and
Koverall = 1E-20
for H2S
Solution
[H+]2 [S2-] Koverall = --- [H2S] Thus,
0.1 [S2-] = 1E-20 (0.1)2
= 1E-19 F
Example 3
Sulfuric acid is a strong acid, and the pKa2 of HSO4- is 1.92. What is the pH of
a 0.100 M NaHSO4 solution?
Solution
The salt is completely ionized in its solution. NaHSO4 Na+ + HSO4
HSO4- = H+ + SO42- Ka2 = 10-1.92 = 0.0120 0.100 - x x, x
Ka2 = x2 / (0.100-x) = 0.0120 [H+] = x
= {-0.120 + (0.0122 + 4*0.00120)1/2} / 2 = 0.0292 M
Thus,
pH = - log 0.0292 = 1.54
Discussion
Is the NaHSO4 salt solution acidic?
Although no concentration is stated, such a solution is acidic because of the acidity of HSO4-.
For H2SO4, pKa2 = 1.92
For H3SO4, pKa1 = 2.12; pKa2 = 7.21; pKa3 = 12.67
Which of the following solutions are acidic, basic, or neutral? Na2SO4
NaH2PO4
Na2HPO4
Na3PO4
NaNO3
Work out the answer please, some of these will appear on the examinations. If the concentration of a salt solution is given, you may be required to evaluate the pH or pOH of the solution.
Example 4
What is the pH of a solution containing 0.500 M NaHSO4 and 0.300 M
Na2SO4?
Solution
The 0.500 M solution of NaHSO4 supplies 0.500 M HSO4- as an acid, and similarly, the solution also contains 0.300 M SO42-.
Using the equation:
pH = pKa - log [salt]/[acid] = 1.92 - log (0.300/0.500) = 2.14
Confidence Building Questions
H2SO3 = 2 H+ + SO3
2-Answer 7.9E-10
Consider...
Koverall = K1 * K2
What is the pH of a 1.0 M H2SO3 solution?
(K1 = 1.2E-2, and K2 = 6.6E-8)
Answer 0.98
Consider...
Only K1 matters in this calculation. Using the quadratic formula yields a pH of 0.98. When approximation is used, you'll get a pH of 0.96
If the pH of a 1.0 M H2SO3 solution is 1.0, what is the sulfite ion
concentration? (K1 = 1.2E-2, and K2 = 6.6E-8)
Answer 7.1E-8
Consider...
Koverall = 7.9E-10 = {[H+]2 [SO32-]} / [H2SO3] = 0.12 [SO32-]/0.9
[SO32-] = ??
This is not an easy question to answer!
AcidBase: Titrations
Skills to develop
Calculate the pH and plot it during a titration of a strong acid by a strong
base.
Calculate the pH and plot it when a weak acid is titrated by a strong base
Titrations
In freshman chemistry, we treat titration this way. A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. Typically, the titrant (the known solution) is added from a buret to a known quantity of the analyte (the unknown solution) until the reaction is complete. Knowing the volume of titrant added allows the determination of the concentration of the unknown. Often, an indicator is used to usually signal the end of the reaction, the endpoint.
In hospital and medical labs, automated titration equipments are used. The following sites have some information regarding the automated titrator:
Metrohm Titration Manipulator.
Karl Fischer Titration Automator
AUTOMATIC TITRATOR INFORMATION
For acid-base titration, a modern lab will usually monitor titration with a pH meter which is interfaced to a computer, so that you will be able to plot the pH or other physical quantities versus the volume that is added.
In this module we simulate this experiment graphically without using chemicals. A program that simulates titrations of strong acids and strong bases is very easy, because the calculation of pH in this experiment is very simple.
Titration Curve
The plot of pH as a function of titrant added is called a titration curve. Let us take a look of a titration process: Example 1
Evaluate [H+] and pH in the titration of 10.0 mL 1.0 M
HCl solution with 1.0 M NaOH solution, and plot the titration curve.
Base
added [H+] pH
0 1.0 0.0
1.0 9/11 0.087 2.0 8/12 0.176 5.0 5/15 0.477 Half-equivalent point 8.0 2/18 0.954 9.0 1/19 1.279 9.3 0.7/19.3 1.440 9.5 0.5/19.5 1.591 9.7 0.3/19.7 1.817 9.8 0.2/19.8 2.0 9.9 0.1/19.9 2.300 9.95 0.05/19.95 2.60 10 H2O=HNaCl neutral salt ++OH- pH = 7
Base
Solution
The amount of acid present = Va*Ca = 10.0 mL * 1.0 mol/1000 mL = 10 mmol (mili-mole)
The amount of base NaOH added = Vb*Cb The amount of acid left = Va*Ca - Vb*Cb The concentration of acid and thus [H+] = [Va*Ca - Vb*Cb] / (Va + Vb)
With the above formulation, we can built a table for various values as shown on the right.
Working to learn
Plot the titration curve on a graph based on the data.
Answer the following questions.
At equivalent point, why is pH=7? What formula is used to calculate pH? Why does pH change rapidly at the equivalent point?
Sketch titration curves when the concentrations of both acids and bases are 0.10, 0.0010 and 0.000010 M? What can you conclude from these sketches?
What are [Na+] and [Cl-] at the following points: initially (before any base is added), half-equivalent point; equivalent point, after 10.5 mL NaOH is added, after 20.0 mL NaOH is added?
Well, when you have acquired the skill to calculate the pH at any point during titration, you may write a simulation program to plot the titration curve. Calculations for strong-acid_strong-base titration are simple, but when weak acid or base are involved, the calculations are somewhat more complicated. However, we are interested in this area and some simulation programs are available on the internet.
Yue-Ling Wong has given a Java interactive titration simulation. His website is rather fun to play with and it is nicely done. In fact, the design of Wong's Java interactive
simulation is very much like the design of the DOS CACT version.
In a titration experiment, the amount to add from the buret depends on the condition at the time. At the start, you may add large amount before observing much pH change, but when the titration is at its end or equivalent point, you would like to ad the titrant slowly. In the simulation, you control the rate of titration.
While trying the Java simulation of titration, please answer the following questions.
Identify the buffer area on the titration curve.
In the titration of weak acids, why does the pH change quickly at the beginning?
Confidence Building Questions
o What are the pH of solutions of 10, 1.0, 0.10, 0.010 and 0.0010 M HCl?
Answer pH=-1, 0, 1, 2, and 3
Consider...
No calculators should be used.
o What are [H+] and the pH at the half equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?
Answer [H+]=0.333; pH=0.477
Consider...
Do not forget the dilution factor.
o What are [Na+] at the half equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?
Answer [H+]=0.333;
Consider...
What about [Cl-]?
o What are [H+] and the pH at the equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?
Answer [H+]=1e-7; pH=7
Consider...
This is only a theoretical value.
o What are [Na+] and [Cl-] at the equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?
Answer [Ca+] = [Cl-] = 0.5
Consider...
o What are [H+] and the pH of 1.0 M acetic acid solution? Ka = 1.8e-5
Answer [H+]=0.00424; pH=2.37
Consider...
The approximation [H+]=(C Ka)1/2 can be used.
o What is the pH of the above solution when half of the acid is neutralized by NaOH in the titration?
Answer pH=4.74
Consider...
Do you know why pH = pKa in this case? At this point, the solution is a very good buffer.
o What is the pH of the end point or equivalent point when 1.0 M acetic acid is titrated by 1.0 M NaOH solution?
Answer [OH-]=1.18e-4; pOH=4.78; pH=9.22
Consider...
Justfy the following formula for this condition. [OH-]=(Kw/Ka*C)1/2 What is the pH of a 0.5 M sodium acetate solution?
o A 0.10 M acetic acid solution is titrated with a 0.10 M NaOH solution to the equivalent point. What is the concentration of the acetate ion?
Answer 0.05 M
Consider...
You have doubled the volume in this case, and the concentration is 0.05 M sodium acetate.
Buffer Solutions
Titration of weak acid by a strong base
Skills to develop
Calculate the pH and plot the
titration curve when a weak acid is titrated by a strong base.
Solve buffer problems.
Solve hydration (hydrolyis> problems.
Prepare a buffer of desirable pH.
Buffer Solutions
Buffer solutions contain a weak acid and its salt or a weak base and its salt. The pH values of these solutions do not change much when a little bit of acid or base is added. The blood is a natural buffer, and so are other body fluids and plant fluids due to mixtures of weak acids and bases present in them.
On this page, we explore the reasons why the pH of buffer solutions resists to change. Buffer solutions are required for many chemical experiments. They are also useful to standardize pH meters. Thus, there are many suppliers of buffer solutions.
Tedia Buffer Solutions.
HF Scientific, Inc pH Buffers.
Sensorex Color coded buffers.
There are also computer programs available to help design and make buffer solutions on the internet. For example:
Buffer Maker using the Henderson-Hasselbalch equation.
SIS Scientific Software buffer maker.
Titration of a Weak Acid by a Strong Base
You have investigated how the pH varies in a strong-acid and strong-base Titration. We illustrate the titration of a weak acid by a strong base using the following examples. During the titration process and before the equivalent point is reached, some acid has been neutralized by the strong base, and the solution contains a weak acid and its salt. The solution acts as a buffer.
Example 1.
What is the pH of a Ca M acid solution whose acid dissociation constant is
Ka?
(What is the pH of a Ca M weak acid before starting the titration?)
Solution
Let HA represent the weak acid, and assume x M of it is ionized. Then, the ionization and equilibrium concentration is
x2 Ka = Ca-x
x2 + Kax - CaKa = 0
-Ka + (Ka2 + 4 CaKa)1/2 x = 2
pH = -log(x)
Discussion
The method has been fully discussed in Weak acids and bases equilibrium. Symbols are used here, but approximations may be applied to numerical problems.
Example 2.
Let us make a buffer solution by mixing Va mL of acid HA and Vs mL of its
salt NaA. For simplicity, let us assume both the acid and the salt solutions have the same concentration C M. What is the pH of the so prepared buffer solution? The acid dissociation constant is Ka.
Solution
After mixing, the concentrations Ca and Cs of the acid HA and its salt NaA respectively are
Ca = C Va / (Va+Vs) Cs = C Vs / (Va+Vs)
Assume x M of the acid is ionized. Then, the ionization and equilibrium of the acid is shown below, but the salt is completely dissociated.
HA = H+ + A -Ca-x x x
NaA = Na+ + A Cs Cs
Common ion [A-] = x+Cs
x(x+Cs) Ka = Ca-x
x2 + (Ka+Cs)x - Ca Ka = 0
-(Ka+Cs) + ((Ka+Cs)2 + 4 Ca Ka)1/2 x = 2
Discussion
The formulas for x and the pH derived above can be used to estimate the pH of any buffer solution, regardless how little salt or acid is used compared to their counter part.
When the ratio Ca / Cs is between 0.1 and 10, the Henderson-Hasselbalch equition is a convenient formula to use.
[H+] [A-] Ka = [HA]
[A-] pKa = pH - log (----) [HA]
The Henderson-Hasselbalch equation is [A-]
pH = pKa + log (----) [HA]
[Cs] + x = pKa + log (----) [Ca] - x
Because when x is insignificant in comparision to Cs, [A-] = Cs and [HA] = Ca
Example 3
Plot the titration curve when a 10.00 mL sample of 1.00 M weak acid HA (Ka = 1.0e-5) is titrated with 1.00 M
NaOH.
Solution
A. Because the concentration is high, we use the approximation [H+] = (CaKa)1/2
= 0.00316 pH = 2.500
Note the sharp increase in pH when 0.1 mL (3 drops) of basic solution is added to the solution.
B. When 0.1 mL NaOH is added, the concentration of salt (Cs), and concentration of acid Ca are:
Cs = 0.1*1.0 M / 10.1 = 0.0099 M Ca = 9.9*1.0 M / 10.1 = 0.98 C. HA = H+ +
A-D. Ca-x x x E.
F. [A-] = x + 0.0099 (= Cs)
Base
G.
H. x (x + 0.0099) I. Ka = --- = 1e-5 J. 0.98 - x
K.
L. x2 + 0.0099 x = 9.8e-6 - 1e5 x M. x2 + (0.0099 + 1e5) x - 9.8e-6 = 0 N.
O. x = (-0.0099 + (0.00992 + 4*0.98*1e-5)1/2) / 2 P. = 0.000907
Q.
R. pH = 3.042
Note that using the Henderson-Hasselbalch equation will not yield the correct solution. Do you know why?
S. When 1.0 mL NaOH is added, concentration of salt, Cs = 1.0*1.0 M / 11.0 = 0.0901 M
Ca = 9.0*1.0 M / 11.0 = 0.818 T. HA = H+ +
A-U. Ca-x x x+0.0901 V.
W. x (x + 0.0901) X. Ka = --- = 1e-5 Y. 0.818 - x
Z.
AA. x = (-0.0901 + (0.09012 + 4*0.818*1e-5)1/2) / 2 BB. = 0.0000907 (any approximation to be made?) CC.
DD. pH = 4.042
Using the Henderson Hasselbalch equation yield 0.0901(=[A-])
pH = pKa + log (---) = 5 - 0.958 = 4.042 (same result) 0.818(=[HA])
EE.When 5.0 mL NaOH is added, concentration of salt, Cs = 5.0*1.0 M / 15.0 = 0.333 M
Ca = 5.0*1.0 M / 15.0 = 0.333 M FF. HA = H+ +
A-GG. Ca-x x x+0.3333 HH.
II. x (x + 0.333)
JJ. Ka = --- = 1.0e-5 KK. 0.333 - x
LL.
MM. x 2 + (0.333-1e-5)x - 0.333*1e-5 = 0 NN.
OO. x = (-0.333 + (0.3332 + 4*3.33e-6)1/2) / 2 PP. = 0.0000010
QQ.
Note: Using the Henderson Hasselbalch equation yield the same result 0.333(=[A-]
pH = pKa + log (----) = 5 + 0.000 = 5.000 0.333(=[HA])
SS. When 10.0 mL NaOH is added, concentration of salt, Cs = 10.0*1.0 M / 20.0 = 0.500 M
Ca = 0.0*1.0 M / 20.0 = 0.000
At the equivalent point, the solution contains 0.500 M of the salt NaA, and the following equilibrium must be considered:
TT. A- + H2O = HA + OH
-UU. Cs-x x x Equilibrium concentrations VV. [HA] [OH-] [H+]
WW. Kb = -- ----XX. [A-] [H+] YY.
ZZ. Kw 1e-14 x 2 AAA. = --- = --- = 1e-9 = ---BBB. Ka 1e-5 Cs-x CCC.
DDD. x = (0.500*1.0e-9)1/2 = 2.26e-5 EEE.
FFF. pOH = -log x = 2.651; pH = 14 - 2.651 = 11.349
Note: The calculation here illustrate the hydration or hydrolysis of the basic salt NaA.
Discussion..
Sketch the titration curve based on the estimates given in this example, and notice the points made along the way.
Confidence Building Question
Plot the titration curve when a 10.00 mL sample of 0.100 M weak acid HA (Ka = 10e-6) is titrated by 0.100 M NaOH solution.
Hint..
Learn it by doing! Perform all the calculations outlined in Example 3, and then you will be an expert for solving buffer and hydration problems.
The compound HF is a weak acid, Ka = 6.7e-4. Calculate Kb and pKb for the
fluoride ion F- in an aqueous solution?
Hint..
The following relationship is handy to use, but make sure you know why Kb = Kw / Ka = ?
